Question

A.) For the reaction

2A(g)+2B(g)⇌C(g)

Kc = 56.8 at a temperature of 341 ∘C .

Calculate the value of Kp.

B.For the reaction

X(g)+3Y(g)⇌2Z(g)

Kp = 2.60×10⁻² at a temperature of 301 ∘C .

Calculate the value of Kc.

Answer 1

relation between Kc and Kp is

Kc = Kp / (RT)^Δn

where

Δn = (Total moles of gas on the products side) - (Total moles of gas on the reactants side)

R = gas constant = 0.0821 L atm / mol-K

T = temperature in kelvins

comming to Part A

T = 341ºC + 273 = 614 K

Kc = 56.8

Δn = 1-4 = -3

now put all these in you equation

56.8 = Kp / (0.0821 x 614)^-3

Kp = 4.4 x 10^-4

Δn = 2-4 = -2

similarly Part B

T = 273 + 301 = 574 K

Kp = 2.60×10⁻²

Kc = 2.60×10⁻² / (0.0821 x 574 )^-2

Kc = 57.74